JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Permutations and Combinations

The sum of all 3-digit numbers less than or equal to 500, that are formed without using the digit "1" and they all are multiple of 11, is ____.

Answer: 7744

Solution

The three-digit multiples of

11

not exceeding

500

are

11 k

for

k = 10, 11, \dots, 45

, and their total is

11 k = 10 \sum 45 k = 11 \cdot \frac{( 10 + 45 ) \times 36}{2} = 11 \times 990 = 10890

Among them, those with a leading digit

1

are

110, 121, 132, \dots, 198

, that is

11 k

for

k = 10, 11, \dots, 18

, whose sum is

11 (10 + 11 + \dots + 18) = 11 \times 126 = 1386

The remaining ones containing the digit

1

are

231, 319, 341, 418

and

451

, which add to

1760

. Removing both groups leaves

10890 - 1386 - 1760 = 7744

so the required sum is

7744

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Integer answer